Let $g(x) = f(x) + ih(x)$. Here, $x$ is real, so the function takes a real number and spits out a complex number.
A book I am reading says that if $\int_{-\infty}^\infty e^{-ix} g(x) dx$ is a real number, then $f(x)$ is even, and $h(x)$ is odd.
Why is this true?
The book further says that this can be used to write$$\int_{-\infty}^\infty e^{-ix} g(x) dx = 2\int_0^\infty e^{-ix}g(x) dx$$
Why is this true?
Note
$$\int_{-\infty}^\infty e^{-ix} g(x) dx= $$ $$\int_{-\infty}^\infty( [\cos(x) f(x) + \sin(x)h(x) ]-i[\sin(x) f(x) -\cos(x)h(x)] )dx$$
If $f(x)$ is even and $h(x)$ odd, the imaginary part of the integrand
$$\sin(x) f(x) -\cos(x)h(x)$$
is odd as a whole and its integration over $(-\infty, \infty)$ vanishes. Hence, the overall integral is real.
Furthermore, the real part of the integrand is even, and its integral can then be written as two times of its integral over $(0, \infty)$.
